tìm số nguyên x để các phân số sau là số nguyên:
a )\(\dfrac{13}{x-1}\)
b ) \(\dfrac{x+3}{x-2}\)
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a: Để D là số nguyên thì \(x-3\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{4;2;8;-2\right\}\)
a, \(\dfrac{6}{2x+1}\Rightarrow2x+1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
2x + 1 | 1 | -1 | 2 | -2 | 3 | -3 | 6 | -6 |
2x | 0 | -2 | 1 | -3 | 2 | -4 | 5 | -7 |
x | 0 | -1 | 1/2 ( loại ) | -3/2 ( loại ) | 1 | -2 | 5/2 ( loại ) | -7/2 ( loại ) |
c, \(\dfrac{x-3}{x-1}=\dfrac{x-1-2}{x-1}=1-\dfrac{2}{x-1}\Rightarrow x-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
x - 1 | 1 | -1 | 2 | -2 |
x | 2 | 0 | 3 | -1 |
tương tự ....
\(\dfrac{13}{x-15}\) là số nguyên khi \(x-15\) là ước của 13
\(x-15\in\left\{\pm1;\pm13\right\}\)
\(\Rightarrow x\in\left\{16;14;26;2\right\}\)
a, \(x-1\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)
x-1 | 1 | -1 | 3 | -3 |
x | 2 | 0 | 4 | -2 |
b, \(2x-1\inƯ\left(-4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
2x-1 | 1 | -1 | 2 | -2 | 4 | -4 |
x | 1 | 0 | loại | loại | loại | loại |
c, \(\dfrac{3\left(x-1\right)+10}{x-1}=3+\dfrac{10}{x-1}\Rightarrow x-1\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
x-1 | 1 | -1 | 2 | -2 | 5 | -5 | 10 | -10 |
x | 2 | 0 | 3 | -1 | 6 | -4 | 11 | -9 |
d, \(\dfrac{4\left(x-3\right)+3}{-\left(x-3\right)}=-4-\dfrac{3}{x+3}\Rightarrow x+3\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)
x+3 | 1 | -1 | 3 | -3 |
x | -2 | -4 | 0 | -6 |
a, `2/(x-1) in ZZ`.
`=> 2 vdots x - 1`
`=> x-1 in Ư(2)`
`=> x - 1 in {+-1, +-2}`.
`=> x - 1 = 1 => x = 2`.
`=> x - 1 = -1 => x = 0`.
`=> x - 1 = -2 => x = -1`.
`=> x - 1 = 2 => x = 3`.
Vậy `x = 2, 0, - 1, 3`.
b, `4/(2x-1) in ZZ`
`=> 4 vdots 2x - 1`.
`=> 2x - 1 in Ư(4)`
Vì `2x vdots 2 => 2x - 1 cancel vdots 2`
`=> 2x - 1 in {+-1}`
`=> 2x - 1 = -1 => x = 0`.
`=> 2x - 1 = 1 => x = 1`
Vậy `x = 0,1`.
c, `(x+3)/(x-1) in ZZ`.
`=> x + 3 vdots x - 1`
`=> x - 1 + 4 vdots x - 1`.
`=> 4 vdots x-1`
`=> x -1 in Ư(4)`
`=> x - 1 in{+-1, +-2, +-4}`
`x - 1 = 1 => x = 2`.
`x - 1 = -1 => x = 0`.
`x - 1 = 2 =>x = 3`.
`x - 1 = -2 => x = -1`.
`x - 1 = 4 => x = 5`.
`x - 1 = -4 => x = -3`.
Vậy `x = 2, 0 , +-1, 5, -3`.
a: ĐKXĐ: x<>-1
Để \(\dfrac{x^3-x^2+2}{x-1}\in Z\) thì \(x^3-x^2+2⋮x-1\)
=>\(x^2\left(x-1\right)+2⋮x-1\)
=>\(2⋮x-1\)
=>\(x-1\in\left\{1;-1;2;-2\right\}\)
=>\(x\in\left\{2;0;3;-1\right\}\)
b: ĐKXĐ: x<>2
Để \(\dfrac{x^3-2x^2+4}{x-2}\in Z\) thì \(x^3-2x^2+4⋮x-2\)
=>\(x^2\left(x-2\right)+4⋮x-2\)
=>\(4⋮x-2\)
=>\(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(x\in\left\{3;1;4;0;6;-2\right\}\)
c: ĐKXĐ: x<>-1/2
Để \(\dfrac{2x^3+x^2+2x+2}{2x+1}\in Z\) thì \(2x^3+x^2+2x+2⋮2x+1\)
=>\(x^2\left(2x+1\right)+\left(2x+1\right)+1⋮2x+1\)
=>\(1⋮2x+1\)
=>\(2x+1\in\left\{1;-1\right\}\)
=>\(2x\in\left\{0;-2\right\}\)
=>\(x\in\left\{0;-1\right\}\)
\(\Leftrightarrow2-x\inƯ\left(11\right)=\left\{-11;-1;1;11\right\}\\ \Leftrightarrow x\in\left\{-9;1;3;13\right\}\)
a)
\(\dfrac{13}{x-1}\in Z\\ \Rightarrow\left(x-1\right)\inƯ\left(13\right)\\ \Rightarrow\left(x-1\right)\in\left\{1;-1;13;-13\right\}\\ \Rightarrow x\in\left\{2;0;14;-12\right\}\)
b)
\(\dfrac{x+3}{x-2}=\dfrac{x-2+5}{x-2}=\dfrac{x-2}{x-2}+\dfrac{5}{x-2}=1+\dfrac{5}{x-2}\\ 1+\dfrac{5}{x-2}\in Z\\ \Rightarrow\dfrac{5}{x-2}\in Z\\ \Rightarrow\left(x-2\right)\inƯ\left(5\right)\\ \Rightarrow\left(x-2\right)\in\left\{1;-1;5;-5\right\}\\ \Rightarrow x\in\left\{3;1;7;-3\right\}\)
tham khảo
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